Three taps A, B and C can fill a tank in 10, 18 and 6 hours, respectively. If B is open all the time and A and C are open for one hour each alternatively, starting with A, the task will be filled in.
A B C 10 18 6 90 9 5 15 First day B+A=9+5=14 Second day B+C=5+15=20 fill tank in two days =14+20=34 similarly work done in four days =68 Now remaining work done by A+B- In fifth day, =68+14 =82 Now remaining work =90-82=8 Remaining work done by B+C=8/20=2/5 Total time =5+2/5=27/5
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